A light beam from the point A(3,10) reflects from the straight line $2\mathrm{x}+\mathrm{y}-6=0$  and then passes through the point B(7,2) . Find the equations of the incident and reflected beams. 

The incident ray will passes through the point  A(3,10) and the image of B(7,12) with respect to the line $2\mathrm{x}+\mathrm{y}-6=0$ 
Suppose that the image of  B(7,2) with respect  to the line $2\mathrm{x}+\mathrm{y}-6=0$  is P(h,k)
It implies 
$\begin{array}{l}\frac{h-7}{2}=\frac{k-2}{1}=-\frac{2\left(2\right(7)+2-6)}{2^2+1^2}\left(\because \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-2\left(a{x}_1+b{y}_1+c\right)}{a^2+b^2}\right)\\ \frac{h-7}{2}=\frac{k-2}{1}=\frac{-2\left(10\right)}{5}\\ \frac{h-7}{2}=\frac{k-2}{1}=-4\end{array}$ 
Hence,$\begin{array}{l}\frac{\mathrm{h}-7}{2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-4 \frac{\mathrm{k}-2}{1}=-4\\ \mathrm{h}-7\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-8 \mathrm{and} \mathrm{k}-2=-4\\ \mathrm{h}=-1\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{k}=-2\end{array}$  
The image is   (-1,2)
Equation of incident line is the equation of the line joining  $\mathrm{A}(3,10)\text{ and }\mathrm{P}(-1,-2)$
It implies 
$\begin{array}{l}\mathrm{y}-10=\frac{-2-10}{-1-3}(\mathrm{x}-3)\\ \mathrm{y}-10=3(\mathrm{x}-3)\\ \mathrm{y}-10=3\mathrm{x}-9\\ 3\mathrm{x}-\mathrm{y}+1=0\end{array}$
Therefore, the equation of the incident ray is  $3\mathrm{x}-\mathrm{y}+1=0$
The reflected ray is passing through the point of intersection of the incident ray and the surface line
The point of intersection of $3\mathrm{x}-\mathrm{y}+1=0$  and $2\mathrm{x}+\mathrm{y}-6=0$ is  (1,4)
Hence the equation of the reflected ray is equation of the line joining  (1,4) and  (7,2)
It implies 
$\begin{array}{l}\mathrm{y}-4=\frac{2-4}{7-1}(\mathrm{x}-1)\\ \mathrm{y}-4=-\frac{1}{3}(\mathrm{x}-1)\\ 3\mathrm{y}-12=-\mathrm{x}+1\\ \mathrm{x}+3\mathrm{y}-13=0\end{array}$
Therefore, the equation of the reflected ray is  $\mathrm{x}+3\mathrm{y}-13=0$